From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

---
 source/know/concept/archimedes-principle/index.md | 26 +++++++++++------------
 1 file changed, 13 insertions(+), 13 deletions(-)

(limited to 'source/know/concept/archimedes-principle/index.md')

diff --git a/source/know/concept/archimedes-principle/index.md b/source/know/concept/archimedes-principle/index.md
index dc02d12..364461f 100644
--- a/source/know/concept/archimedes-principle/index.md
+++ b/source/know/concept/archimedes-principle/index.md
@@ -23,7 +23,7 @@ which has a pressure and thus affects it.
 The right thing to do is treat the entire body as being
 submerged in a fluid with varying properties.
 
-Let us consider a volume $V$ completely submerged in such a fluid.
+Let us consider a volume $$V$$ completely submerged in such a fluid.
 This volume will experience a downward force due to gravity, given by:
 
 $$\begin{aligned}
@@ -31,10 +31,10 @@ $$\begin{aligned}
     = \int_V \va{g} \rho_\mathrm{b} \dd{V}
 \end{aligned}$$
 
-Where $\va{g}$ is the gravitational field,
-and $\rho_\mathrm{b}$ is the density of the body.
-Meanwhile, the pressure $p$ of the surrounding fluid exerts a force
-on the entire surface $S$ of $V$:
+Where $$\va{g}$$ is the gravitational field,
+and $$\rho_\mathrm{b}$$ is the density of the body.
+Meanwhile, the pressure $$p$$ of the surrounding fluid exerts a force
+on the entire surface $$S$$ of $$V$$:
 
 $$\begin{aligned}
     \va{F}_p
@@ -44,7 +44,7 @@ $$\begin{aligned}
 
 Where we have used the divergence theorem.
 Assuming [hydrostatic equilibrium](/know/concept/hydrostatic-pressure/),
-we replace $\nabla p$,
+we replace $$\nabla p$$,
 leading to the definition of the **buoyant force**:
 
 $$\begin{aligned}
@@ -54,7 +54,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-For the body to be at rest, we require $\va{F}_g + \va{F}_p = 0$.
+For the body to be at rest, we require $$\va{F}_g + \va{F}_p = 0$$.
 Concretely, the equilibrium condition is:
 
 $$\begin{aligned}
@@ -64,8 +64,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-It is commonly assumed that $\va{g}$ is constant everywhere, with magnitude $\mathrm{g}$.
-If we also assume that $\rho_\mathrm{f}$ is constant on the "submerged" side,
+It is commonly assumed that $$\va{g}$$ is constant everywhere, with magnitude $$\mathrm{g}$$.
+If we also assume that $$\rho_\mathrm{f}$$ is constant on the "submerged" side,
 and zero on the "non-submerged" side, we find:
 
 $$\begin{aligned}
@@ -73,12 +73,12 @@ $$\begin{aligned}
     = \mathrm{g} (m_\mathrm{b} - m_\mathrm{f})
 \end{aligned}$$
 
-In other words, the mass $m_\mathrm{b}$ of the entire body
-is equal to the mass $m_\mathrm{f}$ of the fluid it displaces.
+In other words, the mass $$m_\mathrm{b}$$ of the entire body
+is equal to the mass $$m_\mathrm{f}$$ of the fluid it displaces.
 This is the best-known version of Archimedes' principle.
 
-Note that if $\rho_\mathrm{b} > \rho_\mathrm{f}$,
-then the displaced mass $m_\mathrm{f} < m_\mathrm{b}$
+Note that if $$\rho_\mathrm{b} > \rho_\mathrm{f}$$,
+then the displaced mass $$m_\mathrm{f} < m_\mathrm{b}$$
 even if the entire body is submerged,
 and the object will therefore continue to sink.
 
-- 
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